3.179 \(\int \frac{(g+h x) \sqrt{a-a \sin (e+f x)}}{\sqrt{c+c \sin (e+f x)}} \, dx\)

Optimal. Leaf size=355 \[ \frac{i a h \cos (e+f x) \text{PolyLog}\left (2,-i e^{i (e+f x)}\right )}{f^2 \sqrt{a-a \sin (e+f x)} \sqrt{c \sin (e+f x)+c}}-\frac{i a h \cos (e+f x) \text{PolyLog}\left (2,i e^{i (e+f x)}\right )}{f^2 \sqrt{a-a \sin (e+f x)} \sqrt{c \sin (e+f x)+c}}-\frac{i a h \cos (e+f x) \text{PolyLog}\left (2,-e^{2 i (e+f x)}\right )}{2 f^2 \sqrt{a-a \sin (e+f x)} \sqrt{c \sin (e+f x)+c}}-\frac{i a (g+h x)^2 \cos (e+f x)}{2 h \sqrt{a-a \sin (e+f x)} \sqrt{c \sin (e+f x)+c}}+\frac{a (g+h x) \log \left (1+e^{2 i (e+f x)}\right ) \cos (e+f x)}{f \sqrt{a-a \sin (e+f x)} \sqrt{c \sin (e+f x)+c}}-\frac{2 i a (g+h x) \cos (e+f x) \tan ^{-1}\left (e^{i (e+f x)}\right )}{f \sqrt{a-a \sin (e+f x)} \sqrt{c \sin (e+f x)+c}} \]

[Out]

((-I/2)*a*(g + h*x)^2*Cos[e + f*x])/(h*Sqrt[a - a*Sin[e + f*x]]*Sqrt[c + c*Sin[e + f*x]]) - ((2*I)*a*(g + h*x)
*ArcTan[E^(I*(e + f*x))]*Cos[e + f*x])/(f*Sqrt[a - a*Sin[e + f*x]]*Sqrt[c + c*Sin[e + f*x]]) + (a*(g + h*x)*Co
s[e + f*x]*Log[1 + E^((2*I)*(e + f*x))])/(f*Sqrt[a - a*Sin[e + f*x]]*Sqrt[c + c*Sin[e + f*x]]) + (I*a*h*Cos[e
+ f*x]*PolyLog[2, (-I)*E^(I*(e + f*x))])/(f^2*Sqrt[a - a*Sin[e + f*x]]*Sqrt[c + c*Sin[e + f*x]]) - (I*a*h*Cos[
e + f*x]*PolyLog[2, I*E^(I*(e + f*x))])/(f^2*Sqrt[a - a*Sin[e + f*x]]*Sqrt[c + c*Sin[e + f*x]]) - ((I/2)*a*h*C
os[e + f*x]*PolyLog[2, -E^((2*I)*(e + f*x))])/(f^2*Sqrt[a - a*Sin[e + f*x]]*Sqrt[c + c*Sin[e + f*x]])

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Rubi [A]  time = 0.511259, antiderivative size = 355, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 9, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.257, Rules used = {4604, 6741, 12, 6742, 4181, 2279, 2391, 3719, 2190} \[ \frac{i a h \cos (e+f x) \text{PolyLog}\left (2,-i e^{i (e+f x)}\right )}{f^2 \sqrt{a-a \sin (e+f x)} \sqrt{c \sin (e+f x)+c}}-\frac{i a h \cos (e+f x) \text{PolyLog}\left (2,i e^{i (e+f x)}\right )}{f^2 \sqrt{a-a \sin (e+f x)} \sqrt{c \sin (e+f x)+c}}-\frac{i a h \cos (e+f x) \text{PolyLog}\left (2,-e^{2 i (e+f x)}\right )}{2 f^2 \sqrt{a-a \sin (e+f x)} \sqrt{c \sin (e+f x)+c}}-\frac{i a (g+h x)^2 \cos (e+f x)}{2 h \sqrt{a-a \sin (e+f x)} \sqrt{c \sin (e+f x)+c}}+\frac{a (g+h x) \log \left (1+e^{2 i (e+f x)}\right ) \cos (e+f x)}{f \sqrt{a-a \sin (e+f x)} \sqrt{c \sin (e+f x)+c}}-\frac{2 i a (g+h x) \cos (e+f x) \tan ^{-1}\left (e^{i (e+f x)}\right )}{f \sqrt{a-a \sin (e+f x)} \sqrt{c \sin (e+f x)+c}} \]

Antiderivative was successfully verified.

[In]

Int[((g + h*x)*Sqrt[a - a*Sin[e + f*x]])/Sqrt[c + c*Sin[e + f*x]],x]

[Out]

((-I/2)*a*(g + h*x)^2*Cos[e + f*x])/(h*Sqrt[a - a*Sin[e + f*x]]*Sqrt[c + c*Sin[e + f*x]]) - ((2*I)*a*(g + h*x)
*ArcTan[E^(I*(e + f*x))]*Cos[e + f*x])/(f*Sqrt[a - a*Sin[e + f*x]]*Sqrt[c + c*Sin[e + f*x]]) + (a*(g + h*x)*Co
s[e + f*x]*Log[1 + E^((2*I)*(e + f*x))])/(f*Sqrt[a - a*Sin[e + f*x]]*Sqrt[c + c*Sin[e + f*x]]) + (I*a*h*Cos[e
+ f*x]*PolyLog[2, (-I)*E^(I*(e + f*x))])/(f^2*Sqrt[a - a*Sin[e + f*x]]*Sqrt[c + c*Sin[e + f*x]]) - (I*a*h*Cos[
e + f*x]*PolyLog[2, I*E^(I*(e + f*x))])/(f^2*Sqrt[a - a*Sin[e + f*x]]*Sqrt[c + c*Sin[e + f*x]]) - ((I/2)*a*h*C
os[e + f*x]*PolyLog[2, -E^((2*I)*(e + f*x))])/(f^2*Sqrt[a - a*Sin[e + f*x]]*Sqrt[c + c*Sin[e + f*x]])

Rule 4604

Int[((g_.) + (h_.)*(x_))^(p_.)*((a_) + (b_.)*Sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*Sin[(e_.) + (f_.)*(x_
)])^(n_), x_Symbol] :> Dist[(a^IntPart[m]*c^IntPart[m]*(a + b*Sin[e + f*x])^FracPart[m]*(c + d*Sin[e + f*x])^F
racPart[m])/Cos[e + f*x]^(2*FracPart[m]), Int[(g + h*x)^p*Cos[e + f*x]^(2*m)*(c + d*Sin[e + f*x])^(n - m), x],
x] /; FreeQ[{a, b, c, d, e, f, g, h}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[p] && IntegerQ
[2*m] && IGeQ[n - m, 0]

Rule 6741

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =!= u]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rule 4181

Int[csc[(e_.) + Pi*(k_.) + (f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c + d*x)^m*ArcTanh[E
^(I*k*Pi)*E^(I*(e + f*x))])/f, x] + (-Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Log[1 - E^(I*k*Pi)*E^(I*(e + f*x))],
x], x] + Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Log[1 + E^(I*k*Pi)*E^(I*(e + f*x))], x], x]) /; FreeQ[{c, d, e,
f}, x] && IntegerQ[2*k] && IGtQ[m, 0]

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 3719

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[(I*(c + d*x)^(m + 1))/(d*(m + 1)), x
] - Dist[2*I, Int[((c + d*x)^m*E^(2*I*(e + f*x)))/(1 + E^(2*I*(e + f*x))), x], x] /; FreeQ[{c, d, e, f}, x] &&
IGtQ[m, 0]

Rule 2190

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(d*m)/(b*f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*Log[1 + (b*(F^(g*(e + f*x)))^n)/a], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rubi steps

\begin{align*} \int \frac{(g+h x) \sqrt{a-a \sin (e+f x)}}{\sqrt{c+c \sin (e+f x)}} \, dx &=\frac{\cos (e+f x) \int (g+h x) \sec (e+f x) (a-a \sin (e+f x)) \, dx}{\sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}}\\ &=\frac{\cos (e+f x) \int a (g+h x) \sec (e+f x) (1-\sin (e+f x)) \, dx}{\sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}}\\ &=\frac{(a \cos (e+f x)) \int (g+h x) \sec (e+f x) (1-\sin (e+f x)) \, dx}{\sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}}\\ &=\frac{(a \cos (e+f x)) \int ((g+h x) \sec (e+f x)-(g+h x) \tan (e+f x)) \, dx}{\sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}}\\ &=\frac{(a \cos (e+f x)) \int (g+h x) \sec (e+f x) \, dx}{\sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}}-\frac{(a \cos (e+f x)) \int (g+h x) \tan (e+f x) \, dx}{\sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}}\\ &=-\frac{i a (g+h x)^2 \cos (e+f x)}{2 h \sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}}-\frac{2 i a (g+h x) \tan ^{-1}\left (e^{i (e+f x)}\right ) \cos (e+f x)}{f \sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}}+\frac{(2 i a \cos (e+f x)) \int \frac{e^{2 i (e+f x)} (g+h x)}{1+e^{2 i (e+f x)}} \, dx}{\sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}}-\frac{(a h \cos (e+f x)) \int \log \left (1-i e^{i (e+f x)}\right ) \, dx}{f \sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}}+\frac{(a h \cos (e+f x)) \int \log \left (1+i e^{i (e+f x)}\right ) \, dx}{f \sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}}\\ &=-\frac{i a (g+h x)^2 \cos (e+f x)}{2 h \sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}}-\frac{2 i a (g+h x) \tan ^{-1}\left (e^{i (e+f x)}\right ) \cos (e+f x)}{f \sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}}+\frac{a (g+h x) \cos (e+f x) \log \left (1+e^{2 i (e+f x)}\right )}{f \sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}}+\frac{(i a h \cos (e+f x)) \operatorname{Subst}\left (\int \frac{\log (1-i x)}{x} \, dx,x,e^{i (e+f x)}\right )}{f^2 \sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}}-\frac{(i a h \cos (e+f x)) \operatorname{Subst}\left (\int \frac{\log (1+i x)}{x} \, dx,x,e^{i (e+f x)}\right )}{f^2 \sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}}-\frac{(a h \cos (e+f x)) \int \log \left (1+e^{2 i (e+f x)}\right ) \, dx}{f \sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}}\\ &=-\frac{i a (g+h x)^2 \cos (e+f x)}{2 h \sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}}-\frac{2 i a (g+h x) \tan ^{-1}\left (e^{i (e+f x)}\right ) \cos (e+f x)}{f \sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}}+\frac{a (g+h x) \cos (e+f x) \log \left (1+e^{2 i (e+f x)}\right )}{f \sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}}+\frac{i a h \cos (e+f x) \text{Li}_2\left (-i e^{i (e+f x)}\right )}{f^2 \sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}}-\frac{i a h \cos (e+f x) \text{Li}_2\left (i e^{i (e+f x)}\right )}{f^2 \sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}}+\frac{(i a h \cos (e+f x)) \operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{2 i (e+f x)}\right )}{2 f^2 \sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}}\\ &=-\frac{i a (g+h x)^2 \cos (e+f x)}{2 h \sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}}-\frac{2 i a (g+h x) \tan ^{-1}\left (e^{i (e+f x)}\right ) \cos (e+f x)}{f \sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}}+\frac{a (g+h x) \cos (e+f x) \log \left (1+e^{2 i (e+f x)}\right )}{f \sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}}+\frac{i a h \cos (e+f x) \text{Li}_2\left (-i e^{i (e+f x)}\right )}{f^2 \sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}}-\frac{i a h \cos (e+f x) \text{Li}_2\left (i e^{i (e+f x)}\right )}{f^2 \sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}}-\frac{i a h \cos (e+f x) \text{Li}_2\left (-e^{2 i (e+f x)}\right )}{2 f^2 \sqrt{a-a \sin (e+f x)} \sqrt{c+c \sin (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 1.24138, size = 154, normalized size = 0.43 \[ \frac{\left (e^{i (e+f x)}+i\right ) \sqrt{a-a \sin (e+f x)} \left (4 h \text{PolyLog}\left (2,-i e^{-i (e+f x)}\right )+f \left (f x (2 g+h x)-4 i (g+h x) \log \left (1+i e^{-i (e+f x)}\right )\right )\right )}{\sqrt{2} f^2 \left (e^{i (e+f x)}-i\right ) \sqrt{-i c e^{-i (e+f x)} \left (e^{i (e+f x)}+i\right )^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[((g + h*x)*Sqrt[a - a*Sin[e + f*x]])/Sqrt[c + c*Sin[e + f*x]],x]

[Out]

((I + E^(I*(e + f*x)))*(f*(f*x*(2*g + h*x) - (4*I)*(g + h*x)*Log[1 + I/E^(I*(e + f*x))]) + 4*h*PolyLog[2, (-I)
/E^(I*(e + f*x))])*Sqrt[a - a*Sin[e + f*x]])/(Sqrt[2]*(-I + E^(I*(e + f*x)))*Sqrt[((-I)*c*(I + E^(I*(e + f*x))
)^2)/E^(I*(e + f*x))]*f^2)

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Maple [F]  time = 0.188, size = 0, normalized size = 0. \begin{align*} \int{(hx+g)\sqrt{a-a\sin \left ( fx+e \right ) }{\frac{1}{\sqrt{c+c\sin \left ( fx+e \right ) }}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((h*x+g)*(a-a*sin(f*x+e))^(1/2)/(c+c*sin(f*x+e))^(1/2),x)

[Out]

int((h*x+g)*(a-a*sin(f*x+e))^(1/2)/(c+c*sin(f*x+e))^(1/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (h x + g\right )} \sqrt{-a \sin \left (f x + e\right ) + a}}{\sqrt{c \sin \left (f x + e\right ) + c}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((h*x+g)*(a-a*sin(f*x+e))^(1/2)/(c+c*sin(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

integrate((h*x + g)*sqrt(-a*sin(f*x + e) + a)/sqrt(c*sin(f*x + e) + c), x)

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Fricas [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: UnboundLocalError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((h*x+g)*(a-a*sin(f*x+e))^(1/2)/(c+c*sin(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

Exception raised: UnboundLocalError

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{- a \left (\sin{\left (e + f x \right )} - 1\right )} \left (g + h x\right )}{\sqrt{c \left (\sin{\left (e + f x \right )} + 1\right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((h*x+g)*(a-a*sin(f*x+e))**(1/2)/(c+c*sin(f*x+e))**(1/2),x)

[Out]

Integral(sqrt(-a*(sin(e + f*x) - 1))*(g + h*x)/sqrt(c*(sin(e + f*x) + 1)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (h x + g\right )} \sqrt{-a \sin \left (f x + e\right ) + a}}{\sqrt{c \sin \left (f x + e\right ) + c}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((h*x+g)*(a-a*sin(f*x+e))^(1/2)/(c+c*sin(f*x+e))^(1/2),x, algorithm="giac")

[Out]

integrate((h*x + g)*sqrt(-a*sin(f*x + e) + a)/sqrt(c*sin(f*x + e) + c), x)